Biomedical Engineering Reference
In-Depth Information
(4, 4, 3, 2,1, 3,1, 2, 3, 4)
when pattern X is introduced in the net, is V = X .
This result is deduced from Equation , applied to
the case of N + L components:
Y = . We can now compute
the energy value associated to X and Y :
1
1
1
  2
ˆ
ˆ
ˆ
2
2
E X
(
)
= −
(
N
+ −
L
2
d
(
X
,
X
))
= −
(6
+ −
4
0)
= −
50
−∆
E
(
V
) =
(
N
+ −
L
2
d
(
V
,
X
))
H
2
2
H
2
1
1
  2
ˆ
X
2
E Y
(
)
= −
(
N
+ −
L
2
d
(
Y X
,
))
= −
(6
+ −
4
2·6)
= −
2
V
S o ,
i f
V
X
t h e n
a n d
H
2
2
ˆ
ˆ
1
d
(
V
,
X
) =
d
(
V
,
X
)
N
, and the next in-
H
H
equality holds:
This result ( E ( X )< E ( Y )) implies that Y is not
stored in the net, since it is not a minimum of the
energy function. So, this technique is able to avoid
the apparition of spurious patterns.
It must be noted that it will only be necessary to
consider N neurons, their weights, and the weights
corresponding to the last L neurons, that remain
fixed, and do not need to be implemented.
L
N
=
N
+ −
L
2
N
≤ +
N
L
ˆ
ˆ
2
d
(
V
,
X
)
≤ + −
N
L
2
H
Therefore
ˆ
ˆ
ˆ
− ∆
2
E
(
V
) = (
N
+ −
L
2
d
(
V
,
X
))
2
max{(
N
L
) , (
2
N
+ −
L
2) } =
2
H
ˆ
2
2
= (
N
+ −
L
2)
< (
N
+
L
)
=
− ∆
2
E X
(
)
SOME REMARkS ON THE
CAPACITy OF THE NET
which demonstrates our statement.
Then, in order to load a pattern X , it will suf-
fice to load its augmented version, which will
be the unique state maximizing the decrease of
energy.
In this example, we can see how this method
works. Consider, at first, that W =0, that is E(V) =0
for all state vector V . Then, in the original model
MREM, the pattern X =(3,3,2,1,4,2) is loaded,
and matrix W is updated. Note that, in this case,
N =6 and L =4. Then, if Y =(4,4,3,2,1,3), we can
compute:
In Mérida et al. (2002), authors find an expression
for the capacity parameter α for MREM model in
terms of the number of neurons N and the number
of possible states for each neuron, L , for the case
in which N is big enough to apply the Central
Limit Theorem ( N ≥ 30):
2
A
B
1
z
2
(4.1)
(
N L
,
)
+
N
NC
1
1
where
E X
(
)
= −
(
N
2
d
(
X
,
X
))
2
= −
6
2
= −
18
2
H
2
1
1
1
2
2
2
E Y
(
)
= −
(
N
2
d
(
Y X
,
))
= −
(6
2·6)
= − −
( 6)
= −
18
,
,
H
1) L
2
2
2
2
4
L
8 N
A
= + +
N
3
(
N
1)
B
=
8(
N
C
=
L
L
2
L
Therefore, Y has been also loaded into the
network, since X is one global minimum of
the energy function, and E ( Y )= E ( X ). With the
original model, Y is a spurious pattern. Let us
apply the technique of augmented patterns to
solve this problem. In this case, the augmented
pat ter ns a re: (3, 3, 2,1, 4, 2,1, 2, 3, 4)
and z α is obtained by imposing the condition
that the maximum allowed error probability in
retrieving patterns is p error . For p error =0.01, we get
z α ≈2.326.
Some facts can be extracted from the above
expression.
X =
a nd
 
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